Đáp án:
$a,$ \(\left[ \begin{array}{l}x=k2\pi\\x=\frac{\pi}5+\frac{k2\pi}{5}\end{array} \right.\) $(k∈Z)$
$b$ \(\left[ \begin{array}{l}x=k2\pi\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{array} \right.\) $(k∈Z)$
$c,$\(\left[ \begin{array}{l}x=\frac{k\pi}{3}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{array} \right.\) $(k∈Z)$
Giải thích các bước giải:
$a,sin3x=sin2x$
⇔\(\left[ \begin{array}{l}3x=2x+k2\pi\\3x=\pi-2x+k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=k2\pi\\x=\frac{\pi}5+\frac{k2\pi}{5}\end{array} \right.\) $(k∈Z)$
$b, sin2x-sinx=0$
⇔$sin2x=sinx$
⇔\(\left[ \begin{array}{l}2x=x+k2\pi\\2x=\pi-x+k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=k2\pi\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{array} \right.\) $(k∈Z)$
$c, sin5x+sinx=0$
⇔$2.sin\frac{5x+x}{2}.cos$ $\frac{5s-x}{2}=0$
⇔$2sin3x.cos2x=0$
⇔\(\left[ \begin{array}{l}sin3x=0\\cos2x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}3x=k\pi\\2x=\frac{\pi}{2}+k\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{k\pi}{3}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{array} \right.\) $(k∈Z)$
